Question

  giúp mình bài 4 với ạ

Answer 1

4a.

$A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{100}}$

$2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}$

$\Rightarrow 2A-A=(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}})-(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{100}})$

$\Rightarrow A=1-\frac{1}{2^{100}}$

Answer 2

4b.

$B=1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{100}}$

$3B=3+1+\frac{1}{3}+...+\frac{1}{3^{99}}$

$\Rightarrow 3B-B=(3+1+\frac{1}{3}+...+\frac{1}{3^{99}})-(1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{100}})$

$\Rightarrow 2B=3-\frac{1}{3^{100}}$

$\Rightarrow B=\frac{3}{2}-\frac{1}{2.3^{100}}$